# $\int_{0}^{\infty}e^{-st}h(t)dt=0 \Rightarrow h(t)=0.$

Suppose $h(t)$ is continuous function and $\int_{0}^{\infty}e^{-st}h(t)dt=0 ~\forall~ s>s_{0}$, then prove that $h(t)=0$.

I know "if a function is continuous, non-negative or non-positive, and its integration is zero, then function must be zero", which is intuitively clear.

But here asked question is beyond my knowledge. Here, exponential function is doing some miracle, but how? Would you like to help me?

• is there any additional condition on $h(t)$ or $s$ ? – r9m Apr 3 '15 at 17:16
• $s$ can be assumed some positive real number, and no additional condition on $h(t)$. – Real Hilbert Apr 3 '15 at 17:20
• This result I needed in order to prove uniqueness of inverse laplace transform. – Real Hilbert Apr 3 '15 at 17:36
• @RealHilbert The down votes aren't surprising considering the criteria used by the community at the moment. Questions must include your thoughts. You basically just said you couldn't get started. I sympathize and I don't even vote according to these criteria, but it as it is. – Git Gud Apr 3 '15 at 17:53
• Since it doesn't work for a particular $s$, it's probably for every $s> s_0$. – Kitegi Apr 3 '15 at 17:57

Here is an argument:

Technical modification. Fix any $s_1 > s_0$, and let

$$H_0 (x) = \int_{0}^{x} e^{-s_1 t}h(t) \, dt \quad \text{and} \quad H(x) = e^{s_1 x}H_0 (x).$$

Then $H_0(x)$ is differentiable and $H_0(x) = o(1)$ as $x \to \infty$. Then for any $s > s_1$, we have

$$\int_{0}^{R} e^{-st} h(t) \, dt = \left[ e^{-(s-s_1)t} H_0(t) \right]_{t=0}^{t=R} + (s - s_1) \int_{0}^{R} e^{-(s-s_1)t} H_0(t) \, dt.$$

Taking $R \to \infty$, we find that

$$\int_{0}^{\infty} e^{-st} H(t) \, dt = \int_{0}^{\infty} e^{-(s-s_1)t} H_0(t) \, dt = 0$$

as well. Moreover, we have an exponential bound $H(x) = o(e^{s_1 x})$. From now on, we work with $H$ instead of $h$.

Main argument. Fix any $s > \max\{s_1, 0\}$. The exponential bound says that $t \mapsto e^{-st}H(t)$ is integrable on $[0, \infty)$. Now for any polynomial $p(x) = \sum a_k x^k$, we have

$$\int_{0}^{\infty} p(e^{-st})e^{-st}H(t) \, dt = \sum a_k \int_{0}^{\infty} e^{-(k+1)st}H(t) \, dt = 0.$$

Now let $\varphi$ be any continuous function supported on a compact subset of $(0, 1)$. By the Stone-Weierstrass theorem, we can approximate $\varphi$ by a polynomial w.r.t. the supremum norm. So

$$\left| \int_{0}^{\infty} (p(e^{-st}) - \varphi(e^{-st})) e^{-st}H(t) \, dt \right| \leq \| p - \varphi\|_{\infty} \int_{0}^{\infty} e^{-st}|H(t)| \, dt.$$

shows that, by taking $p \to \varphi$, we have

$$\int_{0}^{\infty} \varphi(e^{-st})e^{-st}H(t) \, dt = 0$$

as well. Now the integrand of the LHS is zero outside some compact interval, hence by an easy modification of the fundamental theorem of calculus of variation shows that $H \equiv 0$. Consequently, $h \equiv 0$ as well.

• Appreciate your efforts. I will take some time to understand your proof as I am not pure mathematician. – Real Hilbert Apr 3 '15 at 19:02
• (+1) Nice application of some useful machinery (Stone-Weierstrass and FTCoV). By finding an explicit approximation to $\delta(x-\alpha)$ that is made up of functions orthogonal to $h(x)$, some of the machinery is not required. – robjohn Apr 3 '15 at 21:54
• @robjohn, Thank you. I agree that Bernstein polynomial gives a short-cut proof. (Even some versions of Stone-Weierstrass proof exploit this nice family of polynomials, so my proof may be considered indirect.) By the way, depending on what type of integral OP is considering, the technicality I have introduced seems necessary. For example, if $h(x) = e^{x^2} \sin(e^{2x^2})$ then the Laplace transform exists only in improper sense when $s > 0$. – Sangchul Lee Apr 3 '15 at 23:25

The assumption is equivalent to $$\int_0^\infty e^{-sx}\left(e^{-s_0x}h(x)\right)\,\mathrm{d}x=0\tag{1}$$ for all $$s\gt0$$

Since \begin{align} \int_0^\infty\left(e^{-x}-e^{-2x}\right)^n\,\mathrm{d}x &=\int_0^\infty e^{-(n-1)x}\left(1-e^{-x}\right)^n\,\mathrm{d}e^{-x}\\ &=\int_0^1t^{n-1}(1-t)^n\,\mathrm{d}t\\ &=\frac{n!(n-1)!}{(2n)!}\tag{2} \end{align} define $$f_n(x)=n\binom{2n}{n}\left(e^{-x}-e^{-2x}\right)^n\tag{3}$$ Then $$\int_0^\infty f_n(x)\,\mathrm{d}x=1\tag{4}$$ Furthermore, for $$x\gt0$$, $$0\lt4\left(e^{-x}-e^{-2x}\right)\le1\tag{5}$$ and if $$x\ne\log(2)$$, then the right inequality is strict.

According to $$(9)$$ from this answer, $$\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi(n+\frac14)}}\tag{6}$$ Combining $$(3)$$, $$(5)$$, and $$(6)$$, we get that for $$x\gt0$$ $$0\lt f_n(x)\le\sqrt{\frac{n}\pi}\,\left[4\left(e^{-x}-e^{-2x}\right)\right]^n\tag{7}$$ so that outside any neighborhood of $$\log(2)$$, $$f_n(x)$$ eventually tends monotonically to $$0$$.

$$(4)$$ and $$(7)$$ imply that $$\frac{\log(2)}\alpha f_n\left(\frac{\log(2)}\alpha x\right)$$ is an approximation of $$\delta(x-\alpha)$$.

According to the assumption, for any $$n$$ and $$\alpha$$, $$\int_0^\infty\frac{\log(2)}\alpha f_n\left(\frac{\log(2)}\alpha x\right)\left(e^{-s_0x}h(x)\right)\,\mathrm{d}x=0\tag{7}$$ Since $$h(x)$$ is continuous, $$(7)$$ implies that $$h(x)=0$$ for $$x\ge0$$.